A mixture of hydrogen and oxygen has volume 4000 cm3, temperature 300 K, pressure 200 kPa andmass 1.52 g. The ratio of number of moles of hydrogen to number of moles of oxygen in the mixturewill be

To find the ratio of the number of moles of hydrogen to the number of moles of oxygen in the mixture, we can use the ideal gas law equation:

PV = nRT

Where: P = pressure (in kPa) V = volume (in cm3) n = number of moles R = ideal gas constant (8.314 J/(mol·K)) T = temperature (in K)

First, let's convert the volume from cm3 to m3: V = 4000 cm3 = 4000/1000000 m3 = 0.004 m3

Next, let's convert the pressure from kPa to Pa: P = 200 kPa = 200000 Pa

Now, let's rearrange the ideal gas law equation to solve for the number of moles (n): n = PV / RT

Substituting the given values: n = (200000 Pa) * (0.004 m3) / (8.314 J/(mol·K) * 300 K)

Simplifying the equation: n = 0.107 mol

Therefore, the number of moles of hydrogen in the mixture is 0.107 mol.

Since the ratio of hydrogen to oxygen in water is 2:1, the number of moles of oxygen in the mixture will be half of the number of moles of hydrogen.

Therefore, the number of moles of oxygen in the mixture is 0.107 mol / 2 = 0.0535 mol.

The ratio of the number of moles of hydrogen to the number of moles of oxygen in the mixture is approximately 2:1.